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Monday, February 27, 2023

WORD PROBLEMS IN ORGANIC CHEMISTRY

1 Write down the functional isomers of carbonyl compound with molecular formula C3H6O, which isomer will react faster with HCN and why? Explain the mechanism of reaction also. Will the reaction lead to the completion with the conversion of whole reactant into products at reaction condition? If a strong acid is added to the mixture what will be the effect on concentration of the product and why?

 HINTS--- CH3CH2CHO AND CH3COCH3

2 An organic compound X having molecular formula C4H8O gives orange red ppt with 2,4- DNP reagent . It does not reduce Tollens Reagent but gives yellow ppt of iodoform on heating with NaOI. Compound X on reduction with LiAlH4 gives compound Y which undergoes dehydration reaction on heating with conc H2SO4 to form But-2-ene . Identify the compound X and Y and explain the reactions.

  HINT--- X = CH3CH2 CO CH3 y = CH3CH2CH(OH) CH3

3 An organic compound A (C3H4) on hydration in presence of H2SO4 / HgSO4 gives compound B (C3H6O) . Compound B gives white crystalline product D with NaHSO3 . It gives negative Tollens Test and positive iodoform test. On drastic oxidation B gives compound C (C2 H4O2 ) along with formic acid . Identify compound A, B, C , and explain all the reactions.

 HINT--- A-- CH3C=CH

4 An unknown aldehyde A on reacting with alkali gives ɮ--hydroxy aldehyde which loses water to form an unsaturated aldehyde 2- butenal. Another aldehyde B undergoes disproportionation reaction in the presence of conc alkali to form products C and D . C is an aryl alcohol with formula C7H8O. (a) Identify A and B (b) Write the reaction involved. (c) Name the product when B reacts with Zn-Hg and HCl

HINT---- A - CH3CHO B- C6H5CHO

5 An compound X (C2H4O) on oxidation gives Y ( C2H4 O 2) . X undergoes haloform reaction . On treatment with HCN , X forms a product Z which on hydrolysis gives 2-hydroxy propanoic acid. (a) Write down the structure of X and Y . (b) Name the product when X reacts with dil NaOH ( c) Write down the equation for the reaction involved.

HINT-- X- CH3CHO Y- CH3COOH

6 An alkene A molecular formula ( C5H10) on ozonolysis gives a mixture of two compounds B and C . Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling test but forms iodoform . Identify compound A, B, C giving suitable explanation and write the reaction of ozonolysis and iodoform formation either B or C .

  HINT-- CH3 CH = C(CH3) –CH3

7 An organic compound A has characteric odour. On treatment with NaOH it forms compound B and C . Compound B has molecular formula C7H8O which on oxidation gives back A . The compound C is an sodium salt of acid when C is treated with sodalime , it gives an aromatic compound D . Deduce the structure of A, B, C, D . Write the reaction involved.

 HINTS-- A - C6H5CHO

8 A ketone A which undergoes haloform reaction gives compound B on reduction. B on heating with conc H2SO4 gives compound C which forms monoozonide D . The compound D on hydrolysis in presence of Zn dust gives only CH3CHO. Write the structure and IUPAC name of A, B ,C . Write reactions involved.

  HINT-- CH¬3-CO-CH2-CH3

9 Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b) .Compound (b) is reacted with HBr to give C which is isomer of (a) . When (a) is reacted with sodium metal it gives compound (d) , C8H18 WHICH IS DIFFERENT from the comound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equation for all reactions.

10 An aromatic compound A on treatment with aqueous ammonia and heating forms compound B which on heating with Br2 and KOH forms a compound C of molecular formula C6H7N. Write the structure and IUPAC names of the compounds.

11 An organic compound contains 69.77% acarbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollens reagent but forms an addition compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

12 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative , reduce Tollens Reagent and undergoes Cannizzaro reaction. On vigorous oxidation , it gives 1,2-benzenedicarboxylic acid . Identify the compound.

13 An organic compound A molecular formula C8H16O2 was hydrolysed with dilute sulphuric acid to give a carboxylic acid B and an alcohol C . On oxidation of C with chromic acid produced B . C on dehydration gives but-1-ene . Write equations for reactions involved.

14 An organic compound A with molecular formula C8H8O forms an orange red precipitate with 2,4- DNP reagent and gives yellow ppt on heating with iodine in the presence of sodium hydroxide. It neither reduce Tollens or Fehling reagent, nor does it decolourise bromine water or Bayers reagent. On drastic oxidation with chromic acid , it gives a carboxylic acid B having molecular formula C7H6O2. Identify the comound A and B and explain the reaction involved.


Wednesday, February 22, 2023

Important Distinguish between organic pairs

Class XII Chemistry Distinguish between organic pairs

Lucas test

Used to distinguish primary, secondary and tertiary alcohols.

Lucas reagent – equimolar mixture of conc. HCl and anhydrous ZnCl2

Alcohol is treated with Lucas reagent, turbidity due to the formation of insoluble alkyl chloride is observed.

If the turbidity appears immediately, the alcohol is tertiary.

If the turbidity appears in about five minutes, the alcohol is secondary.

A primary alcohol does not react with Lucas reagent at room temperature and hence no turbidity is formed.

Iodoform test

Iodoform test is a test for the CH3-CO group and is characteristic for alcohols and methyl ketones

Reagents- compounds containing CH3-CO or CH3-CH(OH) group, sodium hydroxide and iodine 

Product formed- iodoform,(CHI3) which is a pale yellow in colour


Tollens’ test

Given by aldehydes and formic acid.

Tollen’s reagent- ammoniacal silver nitrate solution 

The aldehydes are oxidised to corresponding carboxylate anion.The reaction occurs in alkaline medium.

A bright silver mirror is produced due to the formation of silver metal.

Also known as Silver mirror test.

Fehling’s test

Fehling reagent comprises of two solutions,

Fehling solution A and Fehling solution B

Fehling solution A is aqueous copper sulphate 

Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt).

On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained

this test is not given by benzaldehyde

Azo dye test

It involves the reaction of any aromatic primary amine with nitrous acid followed by treatment with an alkaline solution of 2-naphthol, when a brilliant yellow , orange or red coloured dye is obtained.

Used to distinguish between aliphatic primary amine and aromatic primary amine.

Hinsberg’s test

Used for distinguishing primary, secondary and tertiary amines.

Hinsberg’s reagent- benzenesulphonyl chloride (C6H5SO2Cl)

In this test amine is shaken with Hinsberg’s reagent in presence of excess of aqueous KOH solution.

A primary amine gives a clear solution which on acidification gives an insoluble N-alkylbenzenesulphonamide.10

A secondary amine gives an insoluble N,N-dialkylbenzenesulphonamide which remains unaffected on addition of acid.

A tertiary amine does not react at all. Therefore, it remains insoluble in the alkaline solution but dissolves on acidification to give a clear solution.

Carbylamine test-

only Primary aliphatic and aromatic amines are treated with chloroform and KOH . they produce isocynide (RNC) compound having foul smells.

Sodium bicarbonate test-

Carboxylic acids react with NaHCO3 with Brisk effeverscene and librates CO2 gas passed lime water becomes milky due to formation of calcium carbonates.futher more passed through it. its milkiness disappears due to formation of calcium bicarbonate 

while phenol being acidic it does not give this test because phenol is weak acid than carboxylic acids(carboxylic acids having most stable resonating structures)

Ferric chloride tests FeCl3.- phenol reacts with FeCl3 ,forms purple/violet coloured complex 

The given organic samples are teated with ferric chlorides If the sample turns to red, green, purple, or blue colouration then it indicates the presence of phenols and its derivatives.

Ferric chloride will not react with any aliphatic alcohol



Differentiate 

1. Phenol and benzoic acid

2. Butan-1-ol and 2-methylpropan-2ol

3. CH3CH2NH2 and(CH3)2NH

4. Ethyl amine and diethyl amine

5. Propan-1-ol and propan-2-ol

6. Methyl ethanoate and ethyl ethanoate

7. Aniline and N-ethyl aniline

8.  N methyl propane-2 amine and N -ethyl-N methyl ethanmine

9. Aniline and benzylamine

10. Ethylamine and Aniline



Monday, January 9, 2023

Type of Hydrogen- Green Hydrogen Brown Hydrogen Grey Hydrogen Blue Hydrogen

Types of Hydrogen on the basis of production

Green Hydrogen

It is produced by electrolysis of water using renewable energy like wind, solar energy. India is front runner for production of green hydrogen.

Brown Hydrogen

It is hydrogen produced by using of coal. Emission gases are released in atmosphere.

Grey Hydrogen

It is hydrogen produced by using natural gas. Emission gases are released into atmosphere.

Blue Hydrogen

In this hydrogen is produced by using natural gas, and emission gases are absorbed and not to be released into atmosphere.

Monday, January 2, 2023

Jawahar Navodaya Vidyalaya online application form for admission class 6

 नवोदय विद्यालय समिति ने सत्र 2024-25 के लिए जवाहर नवोदय विद्यालय चयन परीक्षा 2024-25  वर्ग  6  में प्रवेश के लिए Online आवेदन आमंत्रित किया है |। आवेदन जमा करने के लिए ऑनलाइन पोर्टल 

16 Sept,2024 तक कर सकते। आवेदन  नवोदय विद्यालय समिति  की वेबसाइट पर जाकर मुफ्त में जमा किया जा सकता है। 

 

* Studying in Class 5 in academic year 2024-25 and completing the session from Govt./Govt. recognized school in the same district , where JNV is functioning and to which candidate is seeking admission.

* Studied full academic session in which class and passed classes III& IV from Govt./ Govt. recognized and n born before 01-05-2012 and after 30-04-2014 (Both dates are inclusive). 

JNVST Prospects

Click here

Last date Apply

16 Sept,2024

Examination Date

18.01.2025

Click on the link to fill 

Click Here

Apply online/ registration 

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 Last years question paper- click here to see and download

Registration -  click



योग्यता :

* शैक्षणिक वर्ष 2024-25 में कक्षा 5 में अध्ययनरत

* विद्यालय  सरकार/सरकार के द्वारा मान्यता प्राप्त से सत्र पूरा किया हो । 

* छात्र को उसी जिले आवेदन में आवेदन करना होगा जहां वह वर्त्तमान में अध्ययनरत है तथा  जहां  जवाहर नवोदय विद्यालय संचालित है |

* छात्र वर्ग 3 और  4 का पूर्ण शैक्षणिक सत्र का अध्ययन  सरकार या  सरकार द्वारा मान्यता प्राप्त विद्यालय से  उत्तीर्ण की हो |

आवेदन कैसे करें :


नीचे दिए गए लिंक से आवेदन करने हेतु format download करें और जिस विद्यालय में अभ्यर्थी अध्ययनरत है , उस विद्यालय के प्रधानध्यापक या शिक्षक से सही -सही सूचनाएं अंकित कराएं तदुपरान्त online form Apply कराएं | 


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• Aadhaar details/ Residence certificate issued by competent Government authority. 
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B. Instructions for the candidates:
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    2. The reservations to the OBC candidates shall be implemented as per Central List. The OBC candidates not included in Central list should apply as General Candidate.
    3. Keep the following scanned copies ready before start filling the application.
      1. Candidate's signature. (Size of signature should be between 10-100 kb for photo)
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      4. Certificate signed by parent and candidate. (Size of images should be between 50-300 kb for photo)
  1. For further details please read Prospectus  : Click Here

आवश्‍यक सूचनाऐं

  1. अपलोड करने के लिए प्रमाण-पत्र -- (डाउनलोड करने हेतु यहाँ क्‍लिक करें)
  2. अभ्‍यर्थियों के लिए निर्देश-:
    1. ऑनलाइन आवेदन प्रक्रिया का केवल एक ही चरण है।
    2. अन्‍य पिछड़ा वर्ग के अभ्‍यर्थियों को आरक्षण केवल केन्‍द्रीय सूची के अनुसार दिया जाएगा। केन्‍द्रीय सूची के अन्‍तर्गत नहीं आने वाले अन्‍य पिछड़ा वर्ग के अभ्‍यर्थी कृपया सामान्‍य अभ्‍यर्थी के रूप में आवेदन करें।
    3. आवेदन प्रारम्‍भ करने से पहले कृपया निम्‍नानुसार स्‍केंन कॉपी तैयार रखें।
      1. अभ्‍यर्थी के हस्‍ताक्षर (हस्‍ताक्षर की इमेज का आकार 10-100 के.बी के बीच होना चाहिए।)
      2. अभिभावक के हस्‍ताक्षर (हस्‍ताक्षर की इमेज का आकार 10-100 के.बी के बीच होना चाहिए।)
      3. अभ्‍यर्थी का फोटोग्राफ (फोटोग्राफ का आकार 10-100 के.बी के बीच होना चाहिए।)
      4. अभिभावक तथा अभ्‍यर्थी द्वारा हस्‍ताक्षरित प्रमाण पत्र (हस्‍ताक्षरित प्रमाण पत्र का आकार 50-300 के.बी के बीच होना चाहिए।)
  3. अधिक जानकारी के लिए कृपया विवरणिका देखें। Click Here

Sunday, October 23, 2022

To analyse the given sample of salt for detection of one acidic and one basic radical

 Aim: - To analyse the given sample of salt for detection of one acidic and one basic radical

Observation table-

A) Detection of acidic  radical(anion)

S.N.

EXPERIMENT

OBSERVATION

INFERENCE

1

Salt solution + dil H2SO4

Colourless, colourless gas with brisk effervescence which turn lime

water milky

Group A anion (CO32- may be present)

2.

Confirmative test :

(a)  BaCl2 Test : Salt solution + BaCl2

 

(b)  MgSO4 Test : Salt solution + MgSO4

White ppt of BaCO


White ppt of MgCO3

CO32-Confirmed CO32- Confirmed

 

B. Detection of basic radical(cation)

S.N.

EXPERIMENT

OBSERVATION

INFERENCE

1

Salt + NaOH solution + heat

Colourless gas with

ammonical smell evolves

Zero group is

present NH4+ may be present

2.

Confirmatory test of  NH4+ :-

(a)  put glass rod dipped in conc. HCl acid on the mouth of test tube

(b)  Salt + conc. NaOH + Nessler reagent + Warm

 


Dense white fumes generate around glass rod


Brown solution

 

NH4+ Confirm 


NH4+ Confirm

 

Result: in given sample of salt on acidic radical carbonate (CO32-) and one basic radical ammonium (NH4+) is present.

To analyse the given sample of salt for detection of one acidic and one basic radical

Aim: - To analyse the given sample of salt for detection of one acidic and one basic radical

Observation table-

A) Detection of acidic  radical(anion)

S.N.

EXPERIMENT

OBSERVATION

INFERENCE

1

Salt + Dil. H2SO4 and warm solution

Colourless gas with vinegar like smell

evolves

Acetate (CH3COO-) may be present

2.

Confirmatory test for Acetate:

(a)  Oxalic acid test: take 1 g of salt in test tube and add small amount of oxalic acid + one -two drops of distilled water

 

(b)  Ferric Chloride test: take 1 g of salt in test tube and add 1-2 ml distilled water , shake well and add 1-2 ml of neutral ferric chloride

solution to it

 


Smell of vinegar obtained

 

 

 

Deep red color appears

 

Acetate (CH3COO-) is confirmed

 

 

 

Acetate (CH3COO-) is confirmed

 

Detection of basic radical(cation)

S.N.

EXPERIMENT

OBSERVATION

INFERENCE

1

Salt + NaOH solution + heat

No gas evolve

Zero group is absent

2.

Prepare original solution of salt by dissolving in 20 ml of water

Cleat solution is formed

Water soluble salt is present

3.

Small amount of OS + Dil. HCl

solution

White precipitate is

formed

Group -I is present

Pb2+ may be present

4.

Confirmatory test of Pb2+: dissolve the precipitate in hot water and divide it into three parts

(a)  part -1 + KI solution


(b)  part-2 + potassium chromate

 

(c)  part-3 + few drops of alcohol + dil. H2SO4

 

 

 

 Yellow precipitate 


yellow precipitate is obtained which is soluble in NaOH 


White precipitate is obtained which is

soluble in ammonium acetate

 

 

 

Pb2+ is confirmed 



Pb2+is confirmed

 



Pb2+ is confirmed

Result: in given sample of salt on acidic radical acetate (CH3COO-) and one basic radical lead (Pb2+) is present. 

feature post

CARBON AND ITS COMPOUNDS-Functional Group

1. Name the functional groups present in the following compounds. a)     CH 3  – CO –CH 2 –CH 2  –CH 2 –CH 3 b)    CH...